Problem: As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as <image>, where k is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.

A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number n, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!

Input

The first input line contains an integer n (1 ≤ n ≤ 109).

Output

Print "YES" (without the quotes), if n can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).

Examples

Input

256


Output

YES


Input

512


Output

NO

Note

In the first sample number <image>.

In the second sample number 512 can not be represented as a sum of two triangular numbers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
long long n, m1, m2, k;
long long sum;
int judge = 0;
void judge_(long long x) {
  if (x % 2 == 0) {
    m1 = (x / 2) * (x + 1);
  } else {
    m1 = ((x + 1) / 2) * x;
  }
  m2 = n - m1;
  long long l = 1, r = sqrt(m2 * 2);
  while (r - l >= 0) {
    long long mid = (l + r) / 2;
    if (mid % 2 == 0) {
      sum = (mid / 2) * (mid + 1);
    } else {
      sum = ((mid + 1) / 2) * mid;
    }
    if (sum < m2) {
      l = mid + 1;
    } else if (sum == m2) {
      judge = 1;
      break;
    } else {
      r = mid - 1;
    }
  }
}
int main() {
  scanf("%lld", &n);
  if (n == 0) {
    printf("NO");
    return 0;
  }
  for (long long i = 1; i <= sqrt(2 * n) + 1; i++) {
    judge_(i);
  }
  if (judge == 0)
    printf("NO");
  else
    printf("YES");
  return 0;
}